A block of mass 2,kg moving on a horizontal s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given mass $m = 2\,kg$,initial velocity $u = 4\,ms^{-1}$ at $x = 0.5\,m$.The retarding force is $F = -kx = -12x$.Using Newton's second law,$F = ma

Vedclass · Accepted Answer

$2.0$

Vedclass · Answer

(C) Given mass $m = 2\,kg$,initial velocity $u = 4\,ms^{-1}$ at $x = 0.5\,m$.The retarding force is $F = -kx = -12x$.Using Newton's second law,$F = ma$,so $a = \frac{F}{m} = \frac{-12x}{2} = -6x$.We know $a = v \frac{dv}{dx}$,so $v \frac{dv}{dx} = -6x$.Integrating both sides: $\int_{4}^{v} v \, dv = \int_{0.5}^{1.5} -6x \, dx$.$\left[ \frac{v^2}{2} \right]_{4}^{v} = -6 \left[ \frac{x^2}{2} \right]_{0.5}^{1.5}$.$\frac{v^2 - 16}{2} = -3 [ (1.5)^2 - (0.5)^2 ]$.$\frac{v^2 - 16}{2} = -3 [ 2.25 - 0.25 ] = -3 [ 2 ] = -6$.$v^2 - 16 = -12$.$v^2 = 4$,which gives $v = 2\,ms^{-1}$.

Similar Questions

An object of mass $500 \; g$, initially at rest, is acted upon by a variable force whose $X$-component varies with $X$ as shown in the graph. The velocities of the object at $X = 8 \; m$ and $X = 12 \; m$ are, respectively:

The work done by a force $\vec{F} = (-6x^3 \hat{i}) \, N$ in displacing a particle from $x = 4 \, m$ to $x = -2 \, m$ is ............... $J$.

When a rubber band is stretched by a distance $x$,it exerts a restoring force $F = ax + bx^2$,where $a$ and $b$ are constants. The work done in stretching the rubber band by a distance $L$ from its unstretched position is:

$A$ body moves from $x = 0$ to $x = x_1$ under the influence of a force $F = cx$. The work done during this process is:

$A$ toy car of mass $5 \, kg$ moves up a ramp under the influence of a force $F$ plotted against displacement $x$ as shown in the graph. The maximum height attained is given by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module